Maths for Construction - Desklib

MATHS FOR CONSTRUCTION
Name of Student
Institution Affiliation

Task 1
Scenario 1
A=LW; L=w+3.2
26.5= (w+3.2) w; w2+3.2w-26.5=0
w=−3.2 ± √ 3.22−(4 × 1×−26.5)
2
=4.4 or -10.8
(b)
Lift =k × ρ ×V 2 × A
L = (1/2) d v2 s CL
The angle of attack and CL are related and can be found using a Velocity Relationship Curve
Graph
Scenario 2
a)
65 miles→ 60 seconds
65 miles= (65 ×1760 /0.91) m=104104 m
Speed in m/s=104104/60=1735 m/s
65 miles=1 hour; 100 miles=10/65=1.54 hours

30 miles/gallon
1 gallon→3.78 liters
1 mile→1.61 km
30 miles →(30*1.61) km
Consumption in litters/km
=3.78/ (30*1.61)=0.078liters/km
Fuel consumption for the whole journey
{ 3.78
30 ×1.61 × 100
1.61 }=4.844
b)
L = (1/2) d v2 s CL
L = Lift, which must equal the airplane's weight in pounds
d = density of the air. This will change due to altitude. These values can be found in
a I.C.A.O. Standard Atmosphere Table.
v = velocity of an aircraft expressed in feet per second
s = the wing area of an aircraft in square feet
CL = Coefficient of lift, which is determined by the type of airfoil and angle of attack.
Scenario 3
1)
6th term

nth term=a+ ( n−1 ) d
d=2/3b-b=-1/3b
6th term=b+ (6-1)*-1/3b
=-2/3b
kth term
kth term=a+ (k-1)*-1/3b
20th term=15; 20th term=b-19/3b=-16/3b
−16
3 b=15 ; b=−45
16
Sum of the first 20 terms
Sn= n
2 ( a+ an )= 20
2 (−45
16 +15)
=121.875
nth term=arn−1
2)
20th term=1 × 1
2
19
=1
2
19
Sum to infinity
Sn= a .1(1−rn )
1−r

lim
n → ∞
a(1−rn)
1−r
Since |r|<1, then rn→ 0 as n → ∞
Hence
S∞ = a
1−r = 1
1−1
2
=2
2Log (3x) + Log (18x) = 27
log (3x) 2+log 18x=27
9x2+18x=27
X2+2x-3=0
x=−2 ± √ 22− ( 4 ×1 ×−3 )
2 =1∨−3
3)
2 LOGe (3x) + LOGe (18x) = 9
LOGe (3x) 2+LOGe (18x) = 9
3x2+18x=9
9x2+18x-9=0
X2+2x-1=0

¿−2± √ 22− ( 4 × 1×−1 )
2 =0.414∨−2.414
Cosh(X) + Sinh(X) = 5
ex+ e−x
2 +( ex−e− x
2 )=9
ex+ e−x−ex+ e−x
2 =9; 2 ex
2 =9
ex=9
Let ex be p
p=9
x=ln 9=2.19
Cosh (2Y) - Sinh (2Y) = 3
Cosh (2Y) = 1
2 ( e2 y+ e−2 y ) ; Sinh (2Y) = 1
2 ( e2 y−e−2 y )
( e2 y +e−2 y )− ( e2 y−e−2 y ) =3
( e2 y +e−2 y )− ( e2 y +e−2 y )=3
2 e−2 y−3=0 ; e−2 y=2
3
2 y=ln 2
3 → y= 1
2 ln 2
3